User barinder banwait - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T16:29:52Z http://mathoverflow.net/feeds/user/5744 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/123104/results-and-conjectures-on-bounds-on-degrees-of-isogenies/123755#123755 Answer by Barinder Banwait for Results and conjectures on bounds on degrees of isogenies Barinder Banwait 2013-03-06T13:00:11Z 2013-03-06T13:00:11Z <p>Mein lieber Stefan,</p> <p>I think that these sorts of questions are wide-open. As far as I know, even if you fix the number field, and vary over AVs up to some fixed dimension, then it is <em>not</em> clear that there exists a bound on the possible degrees of your $\psi$ (which I'm guessing is a <em>cyclic</em> isogeny). You in any case have to be careful of complex multiplications; for example, if you have an elliptic curve over $\mathbb{Q}(i)$ with CM by the maximal order, then you can create $\mathbb{Q}(i)$-rational prime-degree isogenies for every prime that splits in $\mathbb{Z}[i]$. But if you exclude these CM isogenies, then I think it is believed that such a bound on cyclic isogeny degrees for AVs exists. </p> <p>But even pushing Mazur's isogeny theorem to number fields is fairly recent stuff; we can now say that, if $K$ is a number field not containing the Hilbert Class Field of an imaginary quadratic field, and GRH is true, then there are only finitely many prime-degree isogenies for elliptic curves over $K$; this goes back to Momose, but with some refinements by Agnès David; see also the work of Larson and Vaintrob. These bounds are, however, quite big. I once computed the bound for $\mathbb{Q}(\sqrt{5})$ by computing all the constants in David's paper "Caractère d'isogénie..." and got something like $10^{120}$!</p> <p>Serre's open image theorem holds for abelian varieties $A/K$ of dimension 2,6 and odd, where $End(A) = \mathbb{Z}$; so there is a bound $C(A,K)$ such that, for all primes $l > C(A,K)$, the mod-$l$ representation is surjective. It is believed that this constant can be made independent of $A$. So, if you fix your base $K$, and vary over all AVs $A/K$ of fixed dimension $d$ (=2,6,or odd) which have no extra endomorphisms, then the prime-degree isogenies are bounded by a constant depending only on $K$. But as you well know, this uniformity conjecture is still open even for elliptic curves over $\mathbb{Q}$!</p> <p>The paper "Expander Graphs, gonality, and variation of Galois Representations" by Ellenberg, Hall, and Kowalski has some interesting results about <em>families</em> of AVs; see theorems 4 and 7, for example. In particular, if you have a family of AVs over a number field $K$, then there is an absolute constant $C$ such that, if $l > C$, then almost all members of the family have "large" mod-$l$ image. And (morally) "large" image means you don't get isogenies or torsion. See their paper for more details. </p> <p>Finally, regarding your subquestion about torsion: François gave a link to a paper of Clark and Xarles where they address the question of bounding torsion primes for <em>certain classes</em> of AVs. In particular, they prove that a "generalised Szpiro conjecture" implies uniform boundedness of torsion for all "Hilbert-Blumenthal" abelian varieties. But it seems that doing this for <em>all</em> abelian varieties of a fixed dimension is very hard. </p> http://mathoverflow.net/questions/123537/which-level-structures-on-elliptic-curves-are-twist-invariant Which level structures on elliptic curves are twist-invariant? Barinder Banwait 2013-03-04T15:03:50Z 2013-03-04T16:23:32Z <p>Let $N \geq 5$ be a prime, and $H$ a subgroup of $GL_2(\mathbb{F}_N)$. As shown in Chapter IV of [DeRap], there is a curve $X_H(N)$, defined over $K_N := \mathbb{Q}(\zeta_N)^{\det H}$, which is a coarse moduli space of elliptic curves with <strong>level-$H$ structure</strong>; that is, given any $L$ an extension of $K_N$, the $L$-points of $X_H(N)$ correspond to $\bar{L}$-isomorphism classes of elliptic curves over $L$ whose associated mod-$N$ representation on $N$-torsion points is contained in (a conjugate of) $H$. (Sometimes this is a <em>fine</em> moduli space, but I don't think this is important for this question). </p> <p>It is certainly <em>not</em> the case that, if $E/L$ has level-$H$ structure at $N$, then so too do all twists of $E$. For example, if $H$ were the subgroup <code>$\left\{\left(\begin{array}{cc}1 &amp; \ast \\ 0 &amp; \ast \end{array}\right)\right\}$</code>, corresponding to $X_1(N)$, then it is not true.</p> <p>But this <em>is</em> true sometimes; e.g., if <code>$H = \left\{\left(\begin{array}{cc}\ast &amp; \ast \\ 0 &amp; \ast \end{array}\right)\right\}$</code>, the Borel subgroup; this follows from the slogan: <strong>if an elliptic curve has a rational cyclic isogeny, then so too do all twists</strong>. A similar argument sows this for $H$ being the normaliser of a split Cartan subgroup, viz <code>$\left\{\left(\begin{array}{cc}\ast &amp; 0 \\ 0 &amp; \ast \end{array}\right)\right\} \bigcup \left\{\left(\begin{array}{cc}0 &amp; \ast \\ \ast &amp; 0 \end{array}\right)\right\}$</code>. I'll say that these level structures are <strong>twist-invariant</strong>. </p> <p>This leads me to ask:</p> <blockquote> <p>Is there a characterisation of the twist-invariant level structures? What is it about a subgroup $H$ that makes it twist-invariant?</p> </blockquote> <p>[DeRap] : P. Deligne, M. Rapoport, Les schemas de modules de courbes elliptiques''.</p> http://mathoverflow.net/questions/113856/what-are-some-consequences-of-the-mumford-tate-conjecture What are some consequences of the Mumford-Tate conjecture? Barinder Banwait 2012-11-19T18:42:58Z 2012-11-21T08:16:35Z <p>Let $A/K$ be an abelian variety over a number field. On the one hand we have the singular cohomology group</p> <p>$$V := H^1(A(\mathbb{C}),\mathbb{Q})$$</p> <p>with respect to some fixed embedding $K \subset \mathbb{C}$, and from this we can construct the <strong>Mumford-Tate group</strong> $G_A \subset Aut_\mathbb{Q}(V)$, a reductive algebraic group over $\mathbb{Q}$, associated to the natural Hodge structure of $V$. </p> <p>On the other hand, for each prime $l$, we have the $l$-adic étale cohomology group</p> <p>$$V_l := H^1(A(\overline{K}),\mathbb{Q}_l)$$ </p> <p>with respect to some fixed algebraic closure $\overline{K}$ of $K$, and from this we can construct the <strong>$l$-adic algebraic monodromy group</strong> $G_{A,l} \subset Aut_{\mathbb{Q}_l}(V_l)$, a reductive algebraic group over $\mathbb{Q}_l$, defined as the Zariski closure of the image of the $G_K$-representation acting continuously on $V_l$ (this latter representation being dual to that on the Tate module). The <strong>comparison isomorphism</strong> <code>$V_l \cong V \otimes_\mathbb{Q} \mathbb{Q}_l$</code> allows us to compare the identity component $G^0_{A,l}$ of $G_{A,l}$ with $G_{A,\mathbb{Q}_l}$ (that is, $G_A$, but viewed as an algebraic group over $\mathbb{Q}_l$), and the Mumford-Tate conjecture predicts that these two groups are the same. </p> <p>Here is my question:</p> <blockquote> <p>What are some consequences of the Mumford-Tate conjecture for the arithmetic of abelian varieties?</p> </blockquote> <p>An example of the sort of thing I have in mind is the following <a href="http://arxiv.org/abs/1111.0624" rel="nofollow">recent result</a> of David Zywina: if $A$ as above is absolutely simple, $K$ is <em>large enough</em>, and the Mumford-Tate conjecture for $A$ holds, then the density of good primes $v$ of $K$ where the reduction $A_v/\mathbb{F}_v$ is also absolutely simple is 1 if and only if the endomorphism ring of $A$ is commutative. (Zywina's result is actually finer than this, see the link.)</p> <p>I guess I'm asking for other places in the literature where results of this flavour are proven to follow from MTC. </p> <p>(A word on my motivation: I have a problem regarding abelian varieties which I think may follow from MTC, but I'm rather stuck proving this. I think my cause would be helped if I had more tools available.)</p> http://mathoverflow.net/questions/109298/example-of-a-diophantine-application-of-an-open-image-theorem/109301#109301 Answer by Barinder Banwait for Example of a diophantine application of an open image theorem Barinder Banwait 2012-10-10T14:52:51Z 2012-10-10T14:52:51Z <p>Serre's open image theorem (on page IV-20 in his book "Abelian $l$-adic representations...) for non-CM elliptic curves $E/K$ is equivalent to the statement that, for almost all $l$ (how large depending on $E$ and $K$), the $l$-adic representation attached to $T_l(E)$ is surjective. Ditto for the mod-$l$ representation. These are nice examples of number theoretic applications. Explicit bounds are also known (work of Hall, Cojocaru and others...). Note that how large $l$ must be is expected to be independent of $E$, and should depend only on $K$. For example, if $K = \mathbb{Q}$, it is hoped that 37 is large enough for any non-CM elliptic curve. </p> <p>It is conjectured that, for $A/K$ any abelian variety for which $End_{\overline{K}}(A) = \mathbb{Z}$, there should be a similar open-image theorem. This is known (by work of Serre) when the dimension of $A$ is 2,6, or odd. In particular, for such an $A/K$, and for sufficiently large $l$, the mod-$l$ representation has image $GSp_{2g}(\mathbb{F}_l)$. This is also nice.</p> <p>(Work of Bogomolov says that the $l$-adic image of $A$ is open (with the $l$-adic topology) in <code>$G_{A,l}(\mathbb{Q}_l)$</code> ; here $G_{A,l}$ is the $l$-adic algebraic monodromy group. See <a href="http://www.martinorr.name/blog/2010/11/27/images-of-galois-representations" rel="nofollow">this blog post of Martin Orr</a> for a discussion of these groups.)</p> http://mathoverflow.net/questions/98060/constructing-an-icosahedral-weight-2-eigenform Constructing an icosahedral weight 2 eigenform? Barinder Banwait 2012-05-26T18:59:06Z 2012-09-06T22:20:15Z <p>(This question is a spin-off from <a href="http://mathoverflow.net/questions/97624/the-first-odd-degree-2-artin-representation-for-which-the-artin-conjecture-was-pr/97637#97637" rel="nofollow">this other question</a>, and is largely inspired by it.)</p> <p>Let $f \in S_1(\Gamma_1(800),\chi)$ be the weight-one "icosahedral" eigenform constructed by Buhler in his thesis [1]:</p> <p>$$f = q - iq^3 - ijq^7 - q^9 + jq^{13} + \cdots.$$</p> <p>Here $i = \sqrt{-1}$ and $j = \frac{1+\sqrt{5}}{2}$; (and if you'd like to see the first 360 Fourier coefficients, then see page 69 of loc. cit). It's "icosahedral", because the corresponding representation $\rho_f : G_\mathbb{Q} \to GL_2(\mathbb{C})$ has projective image isomorphic to $A_5$. (If you're wondering what $\chi$ is, it is the product of the character of order 2 and conductor 4 with the character of order 5 and conductor 25 sending 2 to $\zeta_5$.)</p> <p>Here is my basic question:</p> <blockquote> <p>Can I use this $f$ to construct a <em>weight 2</em> cuspidal eigenform $g$ whose associated mod $\lambda$ representation $\bar{\rho}_{g,\lambda}$ (for $\lambda$ some prime ideal of the coefficient field of $g$) has projective $A_5$ image?</p> </blockquote> <p>Here are some thoughts I've had:</p> <p>An idea I first saw in section 1 of Lecture 1 of Gelbart's article in [2] gives me some hope; pick a weight one Eisenstein series $E$ that is congruent to 1 mod $l$ (some rational prime) and consider the product $fE$, which will be a weight 2 cuspform (but not an eigenform). Applying a lifting lemma of Deligne and Serre might produce an eigenform with the desired property at some $\lambda$ lying above $l$. </p> <p>I say "might", because the lemma I'm looking at on page 163 in [2] is working with primes above 3, and 3 may be the only prime for which this lifting works; (and since I'm chiefly interested in characteristics 7, 19 and 61, the approach may fail).</p> <p>Moreover, if I want the answer as a $q$-expansion, then perhaps this lifting is not explicit enough. </p> <p>(I thought about being more demanding in the question and stipulating the coefficient field of $g$ and the characteristic of $\lambda$, but decided against it...)</p> <p>Finally, I had wanted to ask this question starting with the conductor 133 "tetrahedral" form found by Tate and some of his students, because that came first historically (see the previous question), and I'd then be asking about "tetrahedral" weight 2 forms; but I was unable to write down its $q$-expansion; MAGMA gives a Runtime error when you ask it to compute a basis of the weight one forms at level 133 (though it seems to be fine at smaller levels such as 23 and 47). And since I really like $q$-expansions, I used Buhler's form. </p> <p>[1]: J. Buhler: Icosahedral Galois Representations. LNM 654.</p> <p>[2]: G. Cornell, J. Silverman, G. Stevens (eds): Modular Forms and Fermat's Last Theorem. Springer.</p> http://mathoverflow.net/questions/105983/must-the-j-invariant-of-an-elliptic-curve-with-an-isogeny-be-integral Must the $j$-invariant of an elliptic curve with an isogeny be integral? Barinder Banwait 2012-08-30T18:21:27Z 2012-08-31T16:34:04Z <p>Let $K$ be a quadratic field, and $E/K$ a non-CM elliptic curve with a $K$-rational $p$-isogeny, for $p$ a prime. I would like to say the following:</p> <blockquote> <p>For large enough $p$, the $j$-invariant $j(E)$ must be in $O_K$. </p> </blockquote> <p>I would also like to know exactly how large $p$ must be. (This lower bound may very well not depend on $K$.)</p> <p>The above is indeed true when $K$ is replaced with $\mathbb{Q}$; 37 is definitely large enough; (though the true bound may be as small as 17. EDIT: Actually, it's 19. See the comments.)</p> <p>Corollary 4.3 in Mazur's article [1] gives me hope that the above may be true. Assuming condition A (which I'll describe presently), it says (for my set-up) that the only primes which can divide the denominator of the $j$-invariant are the primes above 2 and 3; (moreover, if 3 doesn't ramify in $K$, then it's only the primes above 2 that need concern us). </p> <p>Condition A is that $J_0(p)$, the jacobian of the modular curve $X_0(p)$, possesses an "optimal quotient" whose Mordell-Weil rank over $K$ is zero.</p> <p>So I guess I'm hoping for two things:</p> <ol> <li><p>That these small primes can be dealt with (i.e. for $p$ large enough, they don't arise in the denominator of $j$); and </p></li> <li><p>Condition A can be removed after all of these years (at least for $p$ large enough).</p></li> </ol> <p>EDIT (after comments from Felipe Voloch and Noam Elkies): Merel's "winding quotient" has rank 0 over $\mathbb{Q}$; but I don't think it will have rank zero over every quadratic field. I also don't know over which quadratic fields the winding quotient has rank zero. </p> <p>Noam Elkies is quite right that the desired result is "vacuously true", since it is known (by work of Momose, Theorem B in [2]) that there are only finitely many $p$ for which there is a $K$-rational $p$-isogeny. However, I'd still like a more direct approach to the lower bound question, not using Momose's much stronger result, if there is one…</p> <p>[1]: Mazur, B. "Rational Isogenies of Prime Degree". Inventiones, 1978.</p> <p>[2]: Momose, F. "Isogenies of Prime Degree over Number Fields". Compositio, 1995</p> http://mathoverflow.net/questions/97052/subgroups-of-gl-2-over-a-finite-field/97054#97054 Answer by Barinder Banwait for Subgroups of GL_2 over a finite field Barinder Banwait 2012-05-15T21:13:41Z 2012-05-15T21:13:41Z <p>At the text-book level, take a look at Lang's Algebra, Chapter XVIII, Section 12, in my version is on page 712. Seems to be pretty thorough. </p> <p>There is also section 2 of Serre's 1972 Propriétés galoisiennes des points d'ordre fini des courbes elliptiques''. </p> http://mathoverflow.net/questions/95294/does-the-following-characterization-of-subgroups-of-gl-2-mathbbf-p-generali Does the following characterization of subgroups of $GL_2(\mathbb{F}_p)$ generalise? Barinder Banwait 2012-04-26T20:16:35Z 2012-04-26T20:16:35Z <p>Let $p$ be a prime number. By a <strong>Cartan subgroup</strong> of $GL_n(\mathbb{F}_p)$ I mean an absolutely semisimple maximal abelian subgroup.</p> <p>When $n=2$, it is well-known* that, for $G \subset GL_n(\mathbb{F}_p)$ of order prime to $p$, either $G$ is contained in a Cartan subgroup, or it is contained in the normalizer of a Cartan subgroup, or its image in $PGL_n(\mathbb{F}_p)$ is isomorphic to $A_4$, $A_5$, or $S_4$.</p> <blockquote> <p>I would like to know if there is a similar result for larger even values of $n$;</p> </blockquote> <p>where a subgroup of order prime to $p$ would either be contained in the normaliser of a Cartan, or its projective image be one in a finite list of groups.</p> <p>*See e.g. section 2 of Serre's "Propriétés galoisiennes..." paper. </p> http://mathoverflow.net/questions/88694/bound-for-the-number-of-rational-points-on-the-modular-curve/88707#88707 Answer by Barinder Banwait for Bound for the number of rational points on the modular curve Barinder Banwait 2012-02-17T12:51:29Z 2012-02-18T13:29:39Z <p>Let me restrict throughout to prime $N \geq 23$. This ensures that, indeed, $|X_0(N)(K)|$ is finite, for any number field $K$. </p> <p>If your question is "For given $N$ and $K$, how big is $|X_0(N)(K)|$?", then I don't really know. Certainly you'll have the two cusps, and possibly some "CM points" coming from CM elliptic curves; e.g., if $K = \mathbb{Q}(i)$, and $N$ splits in $K$. But these cusps and CM points are somehow "trivial". Remember Ogg in "Diophantine Equations and modular forms", 1975; "The conclusion toward which we are tending seems to be that modular curves only have rational points for which there is a reason". </p> <p>If your question is "Given a number field $K$, is there a bound $C(K)$ such that, if $N > C(K)$, then $X_0(N)(K)$ is trivial?", then this question was studied by Momose in 1995 ("Isogenies of Prime Degree over number fields", Compositio Mathematica, 97). He proved (Theorem A in loc.cit.) that there is a bound $C(K)$ such that, if $N > C(K)$, then any noncuspidal point in $X_0(N)(K)$ is one of three kinds (which he calls 'Type 1', 'Type 2', 'Type 3'). He then asks "Under what conditions on $K$ are there only finitely many points of these three types?" For instance, he shows that, for $K$ quadratic and not imaginary quadratic of class number one, then $X_0(N)(K)$ has noncuspidal points for only finitely many $N$. </p> <p>Making this $C(K)$ effective has been done in a paper of Agnès David ("Caractère d’isogénie et critères d’irréductibilité", Théorème II, available on the arXiV). This paper also explains effectively how Momose's Type 1 and 2 points can only occur for finitely many primes (the type II case might require GRH for a really strong Effective Cheboratev Density Theorem, I'm not sure). Type 3 points don't occur if you assume your $K$ does not contain an imaginary quadratic field <em>and</em> its Hilbert Class Field. </p> <p>I was curious how large these bounds were in a specific example like $K=\mathbb{Q}(\sqrt{5})$; using the formulae in David's paper, I got that, for $N > 8 \times 10^{119}$, $X_0(N)(K)$ has only the two cusps.</p> <p>Dmitry Vaintrob's answer to <a href="http://mathoverflow.net/questions/42391/rational-isogenies-of-prime-degree" rel="nofollow">this question</a> may also be useful. I think their Preprint is now on the arXiv. </p> <p>EDIT. Regarding your new question "Is there any bound for $|X_0(N)(\mathbb{Q})|$, where $N$ is an arbitrary positive integer?", the answer is yes.</p> <p>As a rough rule, $X_0(N)(\mathbb{Q})$ contains <em>only</em> the cusps. How many cusps are there? Page 107 of Diamond And Shurman's book contains a nice table, which tells you that the number of cusps is $\sum_{d|N}\phi(gcd(d,\frac{N}{d}))$. </p> <p>If $N \leq 10$, or $N = 12,13,16,18,25$, then the genus of $X_0(N)$ is zero, and hence will have infinitely many rational points. If $N = 11,14,15,17,19,21,27$, then the genus is one; it turns out that these curves have rank zero, and in addition to the cusps, have $3,2,4,2,1,4,1$ more rational points, respectively. If $N = 37,43,67,163$, then, in addition to the cusps, there are $2,1,1,1$ more rational points, respectively. The results in this paragraph were certainly known before Mazur (I'm not quite sure exactly when). </p> <p>Fact. For any other integer $N$, there are only the cusps. </p> <p>This fact was proved for prime $N$ by Mazur. In the introduction to his "Rational Isogenies of Prime Degree" paper, he reduced the fact (for all $N$) to dealing with the cases $169,91,65,39,125$. I believe these cases were subsequently dealt with by work of Kenku and Mestre. </p> http://mathoverflow.net/questions/42391/rational-isogenies-of-prime-degree Rational Isogenies of Prime Degree Barinder Banwait 2010-10-16T15:29:41Z 2012-02-17T21:42:31Z <p>Dear MO Community, </p> <p>Let $N$ be a prime, and let $X_0(N)$ be the classical modular curve over $\mathbb{Q}$. We know ([1]) that, if there are noncuspidal points in $X_0(N)(\mathbb{Q})$, then $N \in$ {${ \mbox{primes } \leq 19}$} $\cup$ {37,43,67,163}. </p> <p>The basic question of this post is:</p> <blockquote> <p>Are there similar lists of primes when $\mathbb{Q}$ is replaced by a number field $K$? That is, if we fix a general number field $K$, can we determine the primes $N$ for which $X_0(N)(K)$ has noncuspidal points?</p> </blockquote> <p>Perhaps in this generality the question is hard, so suppose we restrict from now on to imaginary quadratic $K$. Then [1] gives an approach to the question, but with the following snags:</p> <ol> <li><p>We need to construct an "optimal" quotient of $J_0(N)$, call it $A$, such that $A(K)$ has Mordell-Weil rank 0;</p></li> <li><p>We must restrict ourselves to primes $N$ which are inert in $K$. </p></li> </ol> <p>Actually, I don't think (1) is a big problem; provided $N > 48h(K)^3 + 1$, we can take $A = \widetilde{J}$, the Eisenstein quotient. [Speculation : if we took the "winding quotient" instead, maybe we can lower that bound...]</p> <p>When $N$ splits, then we can construct "CM points" on $X_0(N)(\mathbb{C})$, but usually they will not be defined over $K$, and even if they are, there will only be a handful of them. </p> <blockquote> <p>Question: For which $N$ that splits in $K$ do we have points on $X_0(N)(K)$ that are neither cuspidal nor CM? Is there a way to systematically find these points?</p> </blockquote> <p>By "systematically", I guess I mean something like the 'isogeny character' approach of [1], where the hunt for the $N$s comes down to when certain congruences are satisfied mod $N$. </p> <p>Many thanks. </p> <p>[1]: Mazur, B. "Rational isogenies of prime degree", Inventiones Mathematicae, 1978</p> http://mathoverflow.net/questions/72886/why-does-the-definition-of-modularity-demand-weight-2 Why does the definition of modularity demand weight 2? Barinder Banwait 2011-08-14T20:52:53Z 2011-08-29T21:00:21Z <p>Allow me to quote a definition from Gelbart in "<em>Modular Forms and Fermat's Last Theorem</em>":</p> <p><strong>Definition.</strong> Let $E/\mathbb{Q}$ be an elliptic curve. We say that $E$ is <em>modular</em> if there is some normalised eigenform</p> <p>$$f(z) = \sum_{i=1}^{\infty} \ a_ne^{2\pi inz} \in S_2(\Gamma_0(N),\epsilon),$$</p> <p>for some level $N$ and Nebentypus $\epsilon$, such that</p> <p>$$a_q = q + 1 - \#(E(\mathbb{F}_q))$$</p> <p>for almost all primes $q$. </p> <p>This is the basic question of the post:</p> <blockquote> <p>Why is the weight of $f$ taken to be 2? Can I instead take 3, or 4, or 5, or even 19/2, without disturbing the peace?</p> </blockquote> <p>I am aware of other definitions of modularity, some of which don't mention modular forms at all, but nonetheless I feel that weight 2 lurks beneath all of these.</p> <p>I think one approach would involve differentials, and the construction of Eichler-Shimura, but I'm not so sure. Further, perhaps there are several reasons which fit together to tell a nice story. </p> <p>Is it a corollary of <a href="http://mathoverflow.net/questions/69786/why-is-there-a-weight-2-modular-form-congruent-to-any-modular-form" rel="nofollow">this question</a> that it doesn't matter what the weight is?</p> <p>Finally, can I replace $E$ above with any abelian variety, and ask the same question?</p> http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture Why should I believe the Mordell Conjecture? Barinder Banwait 2011-02-19T20:37:43Z 2011-07-07T23:31:20Z <p>It was Faltings who first proved in 1983 the Mordell conjecture, that a curve of genus 2 or more over a number field has only finitely many rational points.</p> <p>I am interested to know why Mordell and others believed this statement in the first place. What intuition is there that the statement must hold? Without reference to any proof, why should the conjecture 'morally' be true? Supposing one had to give a colloquium (to a general mathematically literate audience) on this, how could one convince them without going into details of heights or étale cohomology?</p> <p>Answers I'm not looking for will be of the form "You fool, because it's been proved already", or even "Read Faltings' proof". </p> http://mathoverflow.net/questions/67188/recovering-hecke-l-series-from-artin-l-functions Recovering Hecke L-series from Artin L-functions Barinder Banwait 2011-06-07T22:40:42Z 2011-06-08T05:17:31Z <p>Let $K$ be a number field, $\chi : C_K \to \mathbb{C}^\ast$ a Hecke character (that is, a character of the idèle class group), and $L(\chi,s)$ the corresponding Hecke $L$-series. I wish to understand how one may construct a Galois extension $G = Gal(L/K)$ and a complex representation $\rho : G \to \mathbb{C}^\ast$ such that $\mathcal{L}(L/K,\rho,s) = L(\chi,s)$, where $\mathcal{L}$ here denotes the Artin L-function. </p> <p>I know how this works when $\chi$ factors through a congruence subgroup mod $\mathfrak{m}$, or equivalently, if $\chi$ is a Dirichlet character mod $\mathfrak{m}$; namely, take $L = K^\mathfrak{m}$, the ray class field mod $\mathfrak{m}$, and use the Artin symbol to turn the given character into a Galois character. </p> <p>But I am worried that trying to do the same thing for general $\chi$, replacing the Artin symbol with the map $\phi_K : C_K \to Gal(K^{ab}/K)$, will not work. Indeed, the Wikipedia article on 'Hecke Character' suggests that only the Dirichlet characters are accounted for by Class Field Theory (see the last paragraph in the section 'Definition using ideals'). This worries me. </p> http://mathoverflow.net/questions/59620/rational-points-ala-chabauty-coleman Rational points à la Chabauty-Coleman Barinder Banwait 2011-03-25T23:52:55Z 2011-03-26T07:56:43Z <p>I have been trying to learn the method of Chabauty and Coleman to find rational points on curves; I have been reading an exposition by McCallum and Poonen which was pointed out to me by Emerton in <a href="http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture" rel="nofollow">this question</a>. </p> <p>Let $X$ be a curve of genus $g$ over $\mathbb{Q}$ with jacobian variety $J$, let $p$ be a prime of good reduction, and let $\overline{J(\mathbb{Q})}$ be the $p$-adic closure of the Mordell-Weil group $J(\mathbb{Q})$ in $J(\mathbb{Q}_p)$. Denote by $r'$ the dimension of the $p$-adic manifold $\overline{J(\mathbb{Q})}$. </p> <p>The main assumption of the approach is that $r' &lt; g$. This is automatic if $r &lt; g$, where $r$ is the rank of $J$, because in general one has $r' \leq r$. This last inequality needn't be equality, "since $\mathbb{Z}$-independent points in log $J(\mathbb{Q})$ need not be $\mathbb{Z}_p$-independent". </p> <blockquote> <p>How do I compute $r'$?</p> </blockquote> <p>I wrote down a toy example, that is, $X : y^2 = x^5 + 17$. Here $r = 2$, and the method might work if $r'$ was 0 or 1, but I don't know how to check this. </p> <p>I suspect that $r' = 2$, in which case the method is not even applicable, and I must think harder, but my question is not about this example, rather the general approach. </p> <blockquote> <p>Is there an example of a curve $X$ with $r = g = 2$ but with $r' = 0$ or 1?</p> </blockquote> http://mathoverflow.net/questions/54255/upper-bounds-for-ranks-of-modular-jacobians Upper bounds for ranks of modular jacobians Barinder Banwait 2011-02-03T22:53:04Z 2011-03-24T04:08:23Z <p>The following question came to me earlier as a "side question"; something I'd like to know, but which is not totally necessary for what I'm thinking about or doing:</p> <p>Consider the genus 32 curve $X_0(389)$, and denote its Jacobian variety as $J_0(389)$.</p> <p>I am interested in finding an upper bound for the Mordell-Weil rank of $J_0(389)(\mathbb{Q}(i))$. </p> <p>After thinking about this for some time, I turned to Google, which threw up [1]. Apparently, assuming the Birch-Swinnerton-Dyer conjecture, there is an absolute constant $C > 0$ such that for all primes $q$ sufficiently large, we have</p> <p>$\mbox{rank } J_0(q)(\mathbb{Q}) \leq C \mbox{ dim} J_0(q)$. </p> <p>[Ideally this equation would be in the center]</p> <p>The point of that paper is to show that $C = 6.5$ will do (existence of $C$ having been proved in an earlier paper by the same authors), but, "assuming [also] the Riemann Hypothesis for automorphic $L$-functions, Iwaniec, Luo and Sarnak have recently proved that one could take $C = \frac{99}{100}$". </p> <p>Now I reckon 389 is "sufficiently large", which means (if you believe those conjectures) an upper bound for the rank over $\mathbb{Q}$ is 31. But this feels like it is way too big, maybe because there are generally so few rational points on modular curves.</p> <blockquote> <p>Does anyone know how to get a better upper bound? Or am I wrong in hoping for a smaller upper bound?</p> </blockquote> <p>Furthermore, since I'm interested in $\mathbb{Q}(i)$-rank, there is the following:</p> <blockquote> <p>What is the biggest the rank can jump by when going from $J_0(389)(\mathbb{Q})$ to $J_0(389)(\mathbb{Q}(i))$?</p> </blockquote> <p>I guess this last question can be asked in greater generality, replacing the Js with any abelian variety $A/\mathbb{Q}$. Can the rank jump be arbitrarily large when passing from $\mathbb{Q}$ to $\mathbb{Q}(i)$? Or is there a bound in terms of the dimension of $A$, say?</p> <p>It's my bedtime now, so I'll pick this thread up in 9 or so hours. </p> <p>[1]: "Explicit Upper Bound for the (Analytic) rank of $J_0(q)$". E. Kowalski, P. Michel. Israel J. Math, 2000. <a href="http://www.math.u-bordeaux1.fr/~kowalski/explicit-rank.pdf" rel="nofollow">Preprint Available here</a></p> http://mathoverflow.net/questions/55085/new-proofs-to-major-theorems-leading-to-new-insights-and-results/55089#55089 Answer by Barinder Banwait for New proofs to major theorems leading to new insights and results? Barinder Banwait 2011-02-11T01:00:15Z 2011-02-11T01:00:15Z <p>The Manin-Mumford Conjecture, first proved by Raynaud in 1983, states that the points on a curve $X$ of genus 2 or more that are torsion when embedded into its Jacobian are finite in number. The bound is also independent of how you embed the curve (that is, it is independent of the "base point"). </p> <p>Ken Ribet reproved this using the notion of an "almost rational torsion point". This is cool because it can lead to explicit versions of Manin-Mumford. The idea is that the set we're interested in is contained in the set of almost rational torsion points, which Ribet proves is finite, together with hyperelliptic branch points if the curve happens to be hyperelliptic. </p> <p>So finding the almost rational torsion on jacobians of curves might help make things explicit. Doing this for modular jacobians, Ribet proved that</p> <p><code>$X_0(N) \cap J_0(N)_{tors} = \{0,\infty\}$</code>, </p> <p>unless $X_0(N)$ is hyperelliptic, in which case you simply add the branch points. </p> <p>To be honest I should say that this theorem was first proved by Matthew Baker (who actually gives two proofs), and independently around the same time by A. Tamagawa. Ribet's approach is similar to Bakers' second approach. For a survey, google "Torsion points on modular curves and Galois theory" to summon a pdf by Ribet and Minhyong Kim.</p> http://mathoverflow.net/questions/50343/what-would-you-want-to-see-at-the-museum-of-mathematics/50347#50347 Answer by Barinder Banwait for What would you want to see at the Museum of Mathematics? Barinder Banwait 2010-12-25T16:34:43Z 2010-12-25T16:34:43Z <p>An exhibit on how cryptography works, and how it keeps online payments and transactions secure. Perhaps a demo or game where kids get to code a message, and other kids have to try to decode it.</p> http://mathoverflow.net/questions/34269/cm-rational-points-on-modular-curves CM rational points on modular curves Barinder Banwait 2010-08-02T15:17:35Z 2010-10-16T15:38:47Z <p>Dear MO Community, </p> <p>I am trying to understand Mazur's 1976 notes "Rational points on Modular Curves" (which can be found in Springer Lecture Notes in Mathematics 601).</p> <p>Let N be a prime number, and let $X_0(N)$ be the usual modular curve over $\mathbb{Q}$. Say that a point on it is 'CM' if the elliptic curve corresponding to the point has Complex Multiplication. I am interested in the following sorts of questions (which are suggested by what happens over $\mathbb{Q}$):</p> <ol> <li><p>Let $K$ be a number field that is not $\mathbb{Q}$. For which N can I construct CM points on $X_0(N)(K)$ that are not defined over $\mathbb{Q}$? Are there finitely many such N?</p></li> <li><p>For such N, are there finitely many such CM points?</p></li> </ol> <p>For example, let $K = \mathbb{Q}(\sqrt{-6})$, and let $R = O_K = \mathbb{Z}[\sqrt{-6}]$. Set $E = \mathbb{C}/R$. Choose $N$ such that $R/NR$ has nontrivial radical, or equivalently such that $R/NR = \mathbb{F}_N[\epsilon]$ for $\epsilon$ some nontrivial element in the radical. Then (E,ker $\epsilon$) determines a point $a_E(N)$ on $X_0(N)$ which "is defined over a subfield of index 2 in the ray class field of $R \otimes \mathbb{Q}$, with conductor equal to the conductor of $R$ (which in my example is 1). Is $a_E(N)$ defined over $K$? What is this index-2 subfield to which Mazur refers?</p> <p>I think my second question is equivalent to asking: For such N, are there finitely many imaginary quadratic orders $R$ such that the aforementioned index 2 subfield is K, and such that R/NR has nontrivial radical?</p> http://mathoverflow.net/questions/40595/why-is-the-eisenstein-quotient-a-quotient-of-the-new-part-of-the-jacobian Why is the Eisenstein quotient a quotient of the new part of the Jacobian? Barinder Banwait 2010-09-30T10:28:04Z 2010-09-30T10:28:04Z <p>Dear MO Community, </p> <p>Let $X = X_0(N)_{/\mathbb{Q}}$, and $J$ its jacobian. Mazur defines the Eisenstein quotient of $J$, denoted $\widetilde{J}$, as</p> <p>$0 \rightarrow \gamma_IJ \rightarrow J \rightarrow \widetilde{J} \rightarrow 0.$</p> <p>Here, $I$ is the Eisenstein ideal of the Hecke algebra $\mathbb{T}$, $\gamma_I$ is the kernel of the map $\mathbb{T} \rightarrow \lim_{\leftarrow_m} \mathbb{T}/I^m$, and $\gamma_IJ$ is the sub-abelian variety generated by the images $\alpha J$ for $\alpha \in \gamma_I$.</p> <p>My question is: why is it actually a quotient of $J^{new}_{/\mathbb{Q}}$, the new part of the Jacobian?</p> <p>Mazur does prove that $\widetilde{J}$ is actually a quotient of $J^- = J/(1 + w)J$; is this the key?</p> <p>References:</p> <p>Mazur, B. "Modular curves and the Eisenstein Ideal", Publications Mathématiques de l'IHES, 1977</p> <p>Mazur, B. "Rational isogenies of prime degree", Inventiones mathematicae, 1978</p> http://mathoverflow.net/questions/23092/class-field-theory-for-imaginary-quadratic-fields Class Field Theory for Imaginary Quadratic Fields Barinder Banwait 2010-04-30T10:38:26Z 2010-05-04T00:00:40Z <p>Let $K$ be a quadratic imaginary field, and E an elliptic curve whose endomorphism ring is isomorphic to the full ring of integers of K. Let j be its j-invariant, and c an integral ideal of K. Consider the following tower:</p> <p>K(j,E[c]) / K(j,h(E[c])) / K(j) / K,</p> <p>where h here is any Weber function on E. (Note that K(j) is the Hilbert class field of K). </p> <p>We know that all these extensions are Galois, and any field has ABELIAN galois group over any smaller field, EXCEPT POSSIBLY THE BIGGEST ONE (namely, K(j,E[c]) / K). </p> <p>Questions:</p> <ol> <li>Does the biggest one have to be abelian? Give a proof or counterexample.</li> </ol> <p>My suspicion: No, it doesn't. I've been trying an example with K = Q($\sqrt{-15}$), E = C/O_K, and c = 3; it just requires me to factorise a quartic polynomial over Q-bar, which SAGE apparently can't do.</p> <ol> <li>What about if I replace E[c] in the above by E_tors, the full torsion group? </li> </ol> http://mathoverflow.net/questions/123537/which-level-structures-on-elliptic-curves-are-twist-invariant/123544#123544 Comment by Barinder Banwait Barinder Banwait 2013-03-05T10:17:42Z 2013-03-05T10:17:42Z Thank you all for your help on this question. http://mathoverflow.net/questions/112685/ordinary-vs-non-ordinary-for-gl2-type-abelian-surfaces-over-q Comment by Barinder Banwait Barinder Banwait 2012-11-17T20:45:13Z 2012-11-17T20:45:13Z Section 7 of Pink's Crelle paper &quot;l-adic algebraic monodromy groups, cocharacters, and the Mumford-Tate conjecture&quot; states a conjecture which predicts that, for <i>any</i> abelian variety over a number field, the set of primes where the variety has ordinary reduction has density 1 &quot;in the potential sense&quot;. As Felipe observed, this is a theorem in dimensions 1 and 2. The extension of the base is also conjectured to be the smallest field over which the monodromy groups are connected. http://mathoverflow.net/questions/109248/on-simple-factors-of-modular-jacobians-endomorphism-ring-and-simplicity-of-mod-p Comment by Barinder Banwait Barinder Banwait 2012-10-19T18:32:49Z 2012-10-19T18:32:49Z @Tommaso: Regarding question 1, I have a more basic question: Given your newform $f$, can you tell from $f$ what the index of the order End($A_f$) is in the maximal order? Is there a criterion for when this index is 1? http://mathoverflow.net/questions/109248/on-simple-factors-of-modular-jacobians-endomorphism-ring-and-simplicity-of-mod-p Comment by Barinder Banwait Barinder Banwait 2012-10-19T17:58:50Z 2012-10-19T17:58:50Z @Fran&#231;ois: Yes, you are quite right. http://mathoverflow.net/questions/109298/example-of-a-diophantine-application-of-an-open-image-theorem/109301#109301 Comment by Barinder Banwait Barinder Banwait 2012-10-11T10:16:36Z 2012-10-11T10:16:36Z @Adam: Oh I see. Maybe the theorem is somehow used in level lowering of Frey curves attached to putative solutions of Diophantine equations? Maybe it's even used in the proof of FLT? http://mathoverflow.net/questions/109248/on-simple-factors-of-modular-jacobians-endomorphism-ring-and-simplicity-of-mod-p Comment by Barinder Banwait Barinder Banwait 2012-10-10T11:31:52Z 2012-10-10T11:31:52Z @Fran&#231;ois: That would be true provided the Mumford-Tate conjecture was known for $A_f$, and the field $K_A^{conn}$ in Zywina's paper was $\mathbb{Q}$. I'm not sure about either of these claims. http://mathoverflow.net/questions/109248/on-simple-factors-of-modular-jacobians-endomorphism-ring-and-simplicity-of-mod-p Comment by Barinder Banwait Barinder Banwait 2012-10-10T00:04:14Z 2012-10-10T00:04:14Z If your $A_f$ is absolutely simple, then this paper of Zywina might help with question 2: <a href="http://www.mast.queensu.ca/~zywina/papers/Splitting.pdf" rel="nofollow">mast.queensu.ca/~zywina/papers/Splitting.pdf</a> But I suspect that $A_f$ are usually not absolutely simple. http://mathoverflow.net/questions/107500/counting-higher-dimensional-abelian-varieties-of-a-given-conductor Comment by Barinder Banwait Barinder Banwait 2012-09-18T18:46:50Z 2012-09-18T18:46:50Z The answer to your first question is YES; it is known as &quot;Shafarevich conjecture for abelian varieties&quot; (though it's not a conjecture anymore). See this blog post of Martin Orr for an overview: <a href="http://www.martinorr.name/blog/2011/09/19/finiteness-theorems-for-abelian-varieties" rel="nofollow">martinorr.name/blog/2011/09/19/&hellip;</a> http://mathoverflow.net/questions/105983/must-the-j-invariant-of-an-elliptic-curve-with-an-isogeny-be-integral/106052#106052 Comment by Barinder Banwait Barinder Banwait 2012-09-07T08:50:24Z 2012-09-07T08:50:24Z Thank you for your answer JSE, this approach is indeed very interesting. I've meanwhile realised that Mazur himself proved the existence of a rank zero quotient over imaginary quadratic fields at primes inert in the field. The approach you outline however would work for real quadratic fields also. Thanks again! http://mathoverflow.net/questions/105983/must-the-j-invariant-of-an-elliptic-curve-with-an-isogeny-be-integral Comment by Barinder Banwait Barinder Banwait 2012-08-30T20:32:08Z 2012-08-30T20:32:08Z Thanks all for your comments so far! I've made an edit. http://mathoverflow.net/questions/97052/subgroups-of-gl-2-over-a-finite-field Comment by Barinder Banwait Barinder Banwait 2012-05-15T21:17:37Z 2012-05-15T21:17:37Z I think the classification is Borel, (Split or nonsplit) Cartan, or the exceptionals'', which are projectively $A_4$, $A_5$, $S_4$. See Lang or Serre for more details. http://mathoverflow.net/questions/95428/is-there-a-linear-combination-of-fibonacci-numbers-which-is-a-non-zero-square-inf Comment by Barinder Banwait Barinder Banwait 2012-04-28T12:09:25Z 2012-04-28T12:09:25Z (Incidentally, a quick computer search on $F_n + 3F_m = x^2$ yields only 19 solutions, letting $F_n$ and $F_m$ vary through the first 500 Fibonaccis). The work of Bugeaud, Mignotte and Siksek reduce $F_n = x^p$ to a Thue equation, I don't know if that will help here... http://mathoverflow.net/questions/95428/is-there-a-linear-combination-of-fibonacci-numbers-which-is-a-non-zero-square-inf Comment by Barinder Banwait Barinder Banwait 2012-04-28T12:05:14Z 2012-04-28T12:05:14Z I guess joro means $\sum_ia_iF_i$, where the $F_i$ vary through the Fibonacci numbers. In the simple example of $n=1$, $a_1=1$, for instance, then the answer is &quot;No&quot;; indeed, there are only 3 perfect squares in Fibonacci 0,1,144. But what about (say) $F_n + 3F_m = x^2$; can this have infinitely many solutions in $(n,m,x)$? I <i>think</i> this is what joro is asking. http://mathoverflow.net/questions/95294/does-the-following-characterization-of-subgroups-of-gl-2-mathbbf-p-generali Comment by Barinder Banwait Barinder Banwait 2012-04-27T10:28:26Z 2012-04-27T10:28:26Z @Will: Thanks for your interesting thoughts, especially about this variety. It does make me feel that one first needs a characterisation over $\mathbb{C}$ (or any alg closed field) before tackling my stated question; and Jim's comments suggest to me that this is probably not possible. http://mathoverflow.net/questions/95294/does-the-following-characterization-of-subgroups-of-gl-2-mathbbf-p-generali Comment by Barinder Banwait Barinder Banwait 2012-04-26T22:13:19Z 2012-04-26T22:13:19Z @Jim: Thanks for your comment. But does my definition of Cartan subgroup really force me into the split case? I didn't think that it did.